# Rudin 3.11

1. Homework Statement
Suppose $a_n > 0$, $s_n =a_1 + ... + a_n$, and $\sum a_n$ diverges,

a) Prove that

$$\sum \frac{a_n}{1+a_n}$$

diverges.

2. Homework Equations

3. The Attempt at a Solution
Comparison with a_n fails miserably.

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
is there a typo? i haven't thought about the problem yet but i notice s_n is nowhere in the problem

is there a typo? i haven't thought about the problem yet but i notice s_n is nowhere in the problem
it is in part b which I have not posted (yet).

it is in part b which I have not posted (yet).
ok cool

there must be a mistake. if $\sum a_i$ diverges then you cannot prove that $\sum \frac{a_n}{1+a_n}$ converges. are you to show that the second series diverges? or is the first series supposed to converge?

Grr. I am two for two today. I am supposed to prove the second series diverges. Sorry. I checked that like three times. I don't know what is wrong with me.

Anyone know why the edit button is missing from my opening post and in fact my second post above also?

Last edited:
Kurdt
Staff Emeritus
Gold Member
Anyone know why the edit button is missing from my opening post and in fact my second post above also?
You can only edit 30 minutes after you post.

Integral
Staff Emeritus
Gold Member
The edit post button disappears within 24hrs.

I will do the edit to correct the initial post.

The edit post button disappears within 24hrs.
I began this thread less than 8 hours ago as you can see from the time stamps.

Is it 24 hours or 30 minutes? I remember it definitely used to be longer than 30 minutes. If that has changed I would recommend changing it back.

Last edited:
ahh typo, no wonder!

recall from rudin, sum(a_k) converges iff for all e > 0 there is N s.t. for all m > n > N |sum(a_k, k = n, ..., m)| < e

i'll be extra terse on purpose because this is a good problem, made me think

a_k > 0 by hypothesis(this is used in the proof of course)
claim sum(a_k/(1 + a_k)) converges => sum(a_k) converges

"proof sketch"
since sum(a_k/(1 + a_k)) converges, there is a number 0 < p < 1 and an integer N such that for all k > N, a_k < p.

we can pick an M s.t. for all m > n > M,

|sum(a_k, k = n, ..., m)| = |a_n + ... + a_m| = |(1 + a_n)a_n/(1 + a_n) + ... + (1 + a_m)a_m/(1 + a_m)| < (1 + p)|a_n/(1 + a_n) + ... + a_m/(1 + a_m)| < and now it's obvious

goodluck!

Last edited:
Kurdt
Staff Emeritus
Gold Member
I began this thread less than 8 hours ago as you can see from the time stamps.

Is it 24 hours or 30 minutes? I remember it definitely used to be longer than 30 minutes. If that has changed I would recommend changing it back.
The time was changed recently due to users abusing the function. See the following thread.

ahh typo, no wonder!

recall from rudin, sum(a_k) converges iff for all e > 0 there is N s.t. for all m > n > N |sum(a_k, k = n, ..., m)| < e

i'll be extra terse on purpose because this is a good problem, made me think

a_k > 0 by hypothesis(this is used in the proof of course)
claim sum(a_k/(1 + a_k)) converges => sum(a_k) converges

"proof sketch"
since sum(a_k/(1 + a_k)) converges, there is a number 0 < p < 1 and an integer N such that for all k > N, a_k < p.

we can pick an M s.t. for all m > n > M,

|sum(a_k, k = n, ..., m)| = |a_n + ... + a_m| = |(1 + a_n)a_n/(1 + a_n) + ... + (1 + a_m)a_m/(1 + a_m)| < (1 + p)|a_n/(1 + a_n) + ... + a_m/(1 + a_m)| < and now it's obvious

goodluck!
I do not know how you are getting your p but I understand your idea. Here is a complete proof.

If the series $$\sum \frac{a_n}{1+a_n}$$ converges, then $a_n$ must be bounded. Otherwise, given any $\epsilon>0[/tex] and natural number N, I could find [itex]n \geq N$ so that $$\frac{a_n}{1+a_n}$$ is between $1-\epsilon$ and $\epsilon$. So, the terms of that series would not go to 0. So, let B bound $a_n$.

Then fix \epsilon \geq 0. Use the assumed convergence of $$\sum \frac{a_n}{1+a_n}$$ and the Cauchy criterion to find N so large that n,m \geq N implies that $$\sum_{k=n}^m \frac{a_k}{1+a_k} \leq \epsilon/(1+B)$$. Then we have that

$$\sum_{k=n}^m a_k = \sum_{k=n}^m a_k (1+a_k)/(1+a_k) \leq (1+B) \sum_{k=n}^m a_k/(1+a_k) < \epsilon$$

So convergence of the second series implies convergence of the first.

Please confirm that this is correct.

i don't understand your arguement for bounding a_n, this doesn't mean it's wrong though
everything else makes sense to me

i did this, sum(a_k/(1 + a_k)) converges, so lim a_k/(1 + a_k) = 0, so there is an N s.t. for all k > N, a_k/(1 + a_k) < 1/2, so a_k < 1/2(1 + a_k) = 1/2 + a_k/2, so a_k/2 < 1/2, ie, a_k < 1 for all k > N, so that's enough, the whole p thing is just because this implies there is a 0 < p < 1 s.t. a_k < p for all k > N(sometimes this is very useful so I just pulled it out incase I would need it later), but yea it's not needed, all you need is a bound

Last edited:
i don't understand your arguement for bounding a_n, this doesn't mean it's wrong though
everything else makes sense to me
In short, I was just saying that if we assume \sum \frac{a_n}{a_n+1} converges then \frac{a_n}{a_n+1} must go to zero and then its pretty clear that a_n must be bounded. Does that make sense? There are lots of ways to show that formally. Yes, all we need is a bound and we are just confusing each other getting a bound in different ways. :) Problem solved.

part d:

$$\sum \frac{a_n}{1+n a_n}$$

and

$$\sum \frac{a_n}{1+n^2 a_n}$$
.

When a_n = 1, the first one converges and the second one diverges. I cannot prove that that always happens though.

part d:

$$\sum \frac{a_n}{1+n a_n}$$

and

$$\sum \frac{a_n}{1+n^2 a_n}$$
.

When a_n = 1, the first one converges and the second one diverges. I cannot prove that that always happens though.
i'm not sure what you mean about a_n = 1, if this is the case then sum(a_n) = sum(1) diverges and so the does the first one i think since it becomes sum(1/(1 + n))

some observations, assuming a_n > 0 always I take it
if sum(a_n) converges, then they both converge by comparison since |a_n/(1 + na_n)| <= |a_n| and |a_n/(1 + n^2a_n| <= |a_n|

for the first, note if a_n = 1/n, then the first and sum(a_n) both diverge, if a_n = 1/n^2 then sum(a_n) and the first both converge, so my gut feeling is that for the first, we have a sum(a_n) converges iff the first converges situation

for the second, things don't look as nice, note setting a_n = 1/n^2 then sum(a_n) and the second both converge. on the other hand, setting a_n = 1/n, we have that sum(a_n) diverges yet the second converges! so I think you can't say too much here, i'd look at more examples

i'd focus on the first and try to prove that if it converges, then so does sum(a_n), I think it's true but don't have time to work on it atm

Last edited:
i'm not sure what you mean about a_n = 1, if this is the case then sum(a_n) = sum(1) diverges and so the does the first one i think since it becomes sum(1/(1 + n))
Sorry, I meant that the first one diverges and the second one converges. Sorry.